Question

Integrate the function: $\sqrt {1-4x^2}$

Answer 1

Let $I=\int \sqrt {1-4x^2}dx=\int \sqrt{(1)^2-(2x)^2}dx$

Let 2x=t $\Rightarrow$ 2dx=dt

∴$I = \frac{1}{2}\sqrt{(1)^2-(t)^2}dt$

It is known that, $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

$\Rightarrow I=\frac{1}{2}\bigg[\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}\sin^{-1}t\bigg]+C$

=$\frac{t}{4}\sqrt{1-t^2}+\frac{1}{4}\sin^{-1}2x+C$

=$\frac{2x}{4}\sqrt{1-4x^2}+\frac{1}{4}\sin^{-1}2x+C$

=$\frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}\sin^{-1}2x+C$